I have few lists and a dictionary and would like to create a pd dataframe.
Could someone help me out, I seem to be missing something:
one simple example bellow:
dict={"a": 1, "b": 3, "c": "text1"}
l1 = [1, 2, 3, 4]
l3 = ["x", "y"]
Using series I would do like this:
df = pd.DataFrame({'col1': pd.Series(l1), 'col2': pd.Series(l3)})
and would have the lists within the df as expected
for dict would do
df = pd.DataFrame(list(dic.items()), columns=['col3', 'col4'])
And would expect this result:
col1 col2 col3 col4
1 x a 1
2 y b 3
3 c text1
4
The problem is like this the first df would be overwritten by the second call of pd.Dataframe
How would I do this to have only one df with 4 columns?
I know one way would be to split the dict in 2 separate lists and just use Series over 4 lists, but I would think there is a better way to do this, out of 2 lists and 1 dict as above to have directly one df with 4 columns.
thanks for the help
you can also use pd.concat to concat two dataframe.
df1 = pd.DataFrame({'col1': pd.Series(l1), 'col2': pd.Series(l3)})
df2 = pd.DataFrame(list(dic.items()), columns=['col3', 'col4'])
df = pd.concat([df1, df2], axis=1)
Why not build each column seperately via dict.keys() and dict.values() instead of using dict.items()
df = pd.DataFrame({
'col1': pd.Series(l1),
'col2': pd.Series(l3),
'col3': pd.Series(dict.keys()),
'col4': pd.Series(dict.values())
})
print(df)
col1 col2 col3 col4
0 1 x a 1
1 2 y b 3
2 3 NaN c text1
3 4 NaN NaN NaN
Alternatively:
column_values = [l1, l3, dict.keys(), dict.values()]
data = {f"col{i}": pd.Series(values) for i, values in enumerate(column_values)}
df = pd.DataFrame(data)
print(df)
col0 col1 col2 col3
0 1 x a 1
1 2 y b 3
2 3 NaN c text1
3 4 NaN NaN NaN
You can unpack zipped values of list generated from d.items() and pass to itertools.zip_longest for add missing values for match by maximum length of list:
#dict is python code word, so used d for variable
d={"a": 1, "b": 3, "c": "text1"}
l1 = [1, 2, 3, 4]
l3 = ["x", "y"]
df = pd.DataFrame(zip_longest(l1, l3, *zip(*d.items()),
fillvalue=np.nan),
columns=['col1','col2','col3','col4'])
print (df)
col1 col2 col3 col4
0 1 x a 1
1 2 y b 3
2 3 NaN c text1
3 4 NaN NaN NaN
Related
I have this sample
col1 result
1 A
1,2,3
2 B
2,3,4
3,4
4 D
1,3,4
3 C
Here's my map variable.
vals_to_replace = {'1':'A', '2':'B', '3':'C' , '4':'D'}
I map this to col1, and only getting some values from the col result, not sure why why single value got mapped only.
Any ideas on how to solve it?
Thanks
Maybe this is what works for you:
import pandas as pd
df = pd.DataFrame({'col1': ['1', '1,2,3', '2', '2,3,4', '3, 4', '4', '1,3,4', '3']})
translation = {'1':'A', '2':'B', '3':'C' , '4':'D'}
df['result'] = df.col1.str.translate(str.maketrans(translation))
print(df)
Result:
col1 result
0 1 A
1 1,2,3 A,B,C
2 2 B
3 2,3,4 B,C,D
4 3, 4 C, D
5 4 D
6 1,3,4 A,C,D
7 3 C
I have a dataframe like this:
d = {'col1': ['a', 'b'], 'col2': [2, 4]}
df = pd.DataFrame(data=d)
df
>> col1 col2
0 a 2
1 b 4
and i want to duplicate the rows by col2 and get a table like this:
>> col1 col2
0 a 2
1 a 2
2 b 4
3 b 4
4 b 4
5 b 4
Thanks to everyone for the help!
Here's my solution using some numpy:
numRows = np.sum(df.col2)
blankSpace = np.zeros(numRows,).astype(str)
d2 = {'col1': blankSpace, 'col2': blankSpace}
df2 = pd.DataFrame(data=d2)
counter = 0
for i in range(df.shape[0]):
letter = df.col1[i]
numRowsForLetter = df.col2[i]
for j in range(numRowsForLetter):
df2.at[counter, 'col1'] = letter
df2.at[counter, 'col2'] = numRowsForLetter
counter += 1
df2 is your output dataframe!
Dataframe:
col1 col2
A 0
A 1
A nan
B 0
B 1
C and so on...
I am trying to change 1 to 0, 0 to 1 and nan stays as such in col2 wherever col1=='A'.
Code so far:
df.loc[(df.col1=='A') & (df.col2==0),'col2'] = 2
df.loc[(df.col1=='A') & (df.col2==1),'col2'] = 0
df.loc[(df.col1=='A') & (df.col2==2),'col2'] = 1
# Hope you understand why I am converting 0 to 2 first then to 1.
# Because if I convert all zeroes to 1 then all 1's will be converted to
# 0 in subsequent conversion.
Unique values in col2 are 0,1 and nan.
Is there a correct/better way of doing this?
Also, is there a way to directly swap these numbers instead of assignment operators?
One solution using Series.where and astype(bool) with ~ (NOT operator) and then back to astype(int). Then use loc with boolean indexing to assign back to DataFrame:
df.loc[df.col1.eq('A'), 'col2'] = df.col2.where(df.col2.isna(),
(~df.col2.astype(bool)).astype(int))
[out]
col1 col2
0 A 1.0
1 A 0.0
2 A NaN
3 B 0.0
4 B 1.0
5 C NaN
You can also try with df.mask():
m=df.col1.eq('A')&df.col2.isna() #condition
df.col2=1-df.col2.mask(m)
print(df)
col1 col2
0 A 1.0
1 A 0.0
2 A NaN
3 B 1.0
4 B 0.0
I am trying to change 1 to 0, 0 to 1 and nan stays as such in col2
wherever col1=='A'.
use np.where
df['col2] = np.where(df['col1'] == 'A', np.where(df['col2'] == 1, 0 , np.where(df['col2'].isnull() == True, df['col2'],1)),df['col2'])
Output
col1 col2
0 A 1.0
1 A 0.0
2 A NaN
3 B 0.0
4 B 1.0
5 C 0.0
In this case, you can also use your own function in combination with apply().
# import pandas
import pandas as pd
# make a sample data
list_of_rows = [
{'col1': A, 'col2': 1},
{'col1': A, 'col2': 0},
{'col1': A, 'col2': None},
{'col1': B, 'col2': 0},
{'col1': B, 'col2': 1},
{'col1': B, 'col2': None},
]
# make a pandas data frame
df = pd.DataFrame(list_of_rows)
# define a function
def change_values(row):
if row['col2'] == 0:
return 1
if row['col2'] == 1:
return 0
return row['col2']
# apply function to dataframe
df['col2'] = df.apply(lambda row: change_values(row), axis=1)
How do I take a normal data frame, like the following:
d = {'col1': [1, 2], 'col2': [3, 4]}
df = pd.DataFrame(data=d)
df
col1 col2
0 1 3
1 2 4
and produce a dataframe where the column name is added to the cell in the frame, like the following:
d = {'col1': ['col1=1', 'col1=2'], 'col2': ['col2=3', 'col2=4']}
df = pd.DataFrame(data=d)
df
col1 col2
0 col1=1 col2=3
1 col1=2 col2=4
Any help is appreciated.
Make a new DataFrame containing the col*= strings, then add it to the original df with its values converted to strings. You get the desired result because addition concatenates strings:
>>> pd.DataFrame({col:str(col)+'=' for col in df}, index=df.index) + df.astype(str)
col1 col2
0 col1=1 col2=3
1 col1=2 col2=4
You can use apply to set column name in cells and then join them with '=' and the values.
df.apply(lambda x: x.index+'=', axis=1)+df.astype(str)
Out[168]:
col1 col2
0 col1=1 col2=3
1 col1=2 col2=4
You can try this
df.ne(0).mul(df.columns)+'='+df.astype(str)
Out[1118]:
col1 col2
0 col1=1 col2=3
1 col1=2 col2=4
Given the following dataframe:
import pandas as pd
df = pd.DataFrame({'COL1': ['A', np.nan,'A'],
'COL2' : [np.nan,'A','A']})
df
COL1 COL2
0 A NaN
1 NaN A
2 A A
I would like to create a column ('COL3') that uses the value from COL1 per row unless that value is null (or NaN). If the value is null (or NaN), I'd like for it to use the value from COL2.
The desired result is:
COL1 COL2 COL3
0 A NaN A
1 NaN A A
2 A A A
Thanks in advance!
In [8]: df
Out[8]:
COL1 COL2
0 A NaN
1 NaN B
2 A B
In [9]: df["COL3"] = df["COL1"].fillna(df["COL2"])
In [10]: df
Out[10]:
COL1 COL2 COL3
0 A NaN A
1 NaN B B
2 A B A
You can use np.where to conditionally set column values.
df = df.assign(COL3=np.where(df.COL1.isnull(), df.COL2, df.COL1))
>>> df
COL1 COL2 COL3
0 A NaN A
1 NaN A A
2 A A A
If you don't mind mutating the values in COL2, you can update them directly to get your desired result.
df = pd.DataFrame({'COL1': ['A', np.nan,'A'],
'COL2' : [np.nan,'B','B']})
>>> df
COL1 COL2
0 A NaN
1 NaN B
2 A B
df.COL2.update(df.COL1)
>>> df
COL1 COL2
0 A A
1 NaN B
2 A A
Using .combine_first, which gives precedence to non-null values in the Series or DataFrame calling it:
import pandas as pd
import numpy as np
df = pd.DataFrame({'COL1': ['A', np.nan,'A'],
'COL2' : [np.nan,'B','B']})
df['COL3'] = df.COL1.combine_first(df.COL2)
Output:
COL1 COL2 COL3
0 A NaN A
1 NaN B B
2 A B A
If we mod your df slightly then you will see that this works and in fact will work for any number of columns so long as there is a single valid value:
In [5]:
df = pd.DataFrame({'COL1': ['B', np.nan,'B'],
'COL2' : [np.nan,'A','A']})
df
Out[5]:
COL1 COL2
0 B NaN
1 NaN A
2 B A
In [6]:
df.apply(lambda x: x[x.first_valid_index()], axis=1)
Out[6]:
0 B
1 A
2 B
dtype: object
first_valid_index will return the index value (in this case column) that contains the first non-NaN value:
In [7]:
df.apply(lambda x: x.first_valid_index(), axis=1)
Out[7]:
0 COL1
1 COL2
2 COL1
dtype: object
So we can use this to index into the series
You can also use mask which replaces the values where COL1 is NaN by column COL2:
In [8]: df.assign(COL3=df['COL1'].mask(df['COL1'].isna(), df['COL2']))
Out[8]:
COL1 COL2 COL3
0 A NaN A
1 NaN A A
2 A A A